Ignore imaginary roots in sympy

Question

I’m using sympy to solve a polynomial:

x = Symbol('x')
y = solve(int(row["scaleA"])*x**3 + int(row["scaleB"])*x**2 + int(row["scaleC"])*x + int(row["scaleD"]), x)

y is a list of possible solutions. However, I need to ignore the imaginary ones and only use the real solutions. Also, I would like the solution as a value not an expression. Right now it looks like:

[-2/3 - 55**(1/3)*(-1/2 - sqrt(3)*I/2)/3, -2/3 - 55**(1/3)*(-1/2 + sqrt(3)*I/2)/3, -55**(1/3)/3 - 2/3]

I need the last expression’s value (-2.22756). Are there functions in sympy to simplify this?

Asked By: Kreuzade

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Source

Answer 1

I managed to simply ignore solutions containing the character "I" and used .evalf() to evaluate the expression. The code is now:

    x = Symbol('x')
    y = solve(int(row["scaleA"])*x**3 + int(row["scaleB"])*x**2 + int(row["scaleC"])*x + int(row["scaleD"]), x)
    for root in y:
        if "I" not in str(root):
            print("This One:" + str(root.evalf()))

Answered By: Kreuzade

Answer 2

As Krastonov had mentioned mpmath provided an easier method:

y = polyroots([int(row["scaleA"]), int(row["scaleB"]), int(row["scaleC"]), int(row["scaleD"])-value])
for root in y:
   if "j" not in str(root):
       value = root

Answered By: Kreuzade

Answer 3

If you set x to be real, SymPy will only give you the real solutions

x = Symbol('x', real=True)
solve(..., x)

Answered By: asmeurer

Answer 4

solve() doesn’t have a consistent output for various types of solutions, please use solveset(Eq,x,domain=S.Reals) :

 from sympy import ImageSet, S 
 x = Symbol('x')
 y = solveset(int(row["scaleA"])*x**3 + int(row["scaleB"])*x**2+int(row["scaleC"])*x + int(row["scaleD"]), x, domain=S.Reals)

http://docs.sympy.org/latest/modules/solvers/solveset.html

Answered By: PyRick

Answer 5

This is exactly the sort of thing that real_roots is made for and is especially applicable to your case where the coefficients are integers:

x = Symbol('x')
eq = int(row["scaleA"])*x**3 + int(row["scaleB"])*x**2 + int(row["scaleC"])*x + int(row["scaleD"])
y = real_roots(eq, x)  # gives [CRootOf(...), ...]

The value of CRootOf instances can be evaluated to whatever precision you need and should not contain any imaginary part. For example,

>>> [i.n(12) for i in real_roots(3*x**3 - 2*x**2 + 7*x - 9, x)]
[1.07951904858]

Note: As I recall, solve will send back roots that it wasn’t able to confirm met the assumptions (i.e. if they weren’t found to be false for the assumption then they are returned). Also, if you want more consistent output from solve, @PyRick, set the flag dict=True.

Answered By: smichr

Answer 6

To get rid of complex numbers with just SymPy (without mpmath) use solve with expand_complex or realroots as mentioned above by @smichr.

from sympy import *
x = var("x")

# With symbol p:
p = var("p", positive=True)
# Without symbol p:
p = 1

a, b, c, d = 1, 15, 10, 1
term = p * (a*x**3 + b*x**2 + c*x + d)

pprint("nSolution 1:")
sol = solve(term, x)
for x in sol:
    pprint(x)
    tmp = expand_complex(x)
    pprint(tmp)
    pprint(N(tmp))

pprint("nSolution 2:")
sol = real_roots(term, x)
for x in sol:
    pprint(N(x))

With symbols, use solve with expand_complex.

Answered By: kassbohm

Ignore imaginary roots in sympy

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